--- title: "Insertion loss method and microstrip LPF" subject: "RF and Microwave Engineering" chapter: 6 key: "semester_7/rf_and_microwave_engineering:insertion-loss-lpf-microstrip" source: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/insertion-loss-lpf-microstrip site: BCT Engineering --- # Insertion loss method and microstrip LPF RF and Microwave Engineering · Chapter 6: RF Design Practices - Asked in: 2073 Magh, 8+2 marks - Video: [Design Steps for RF filter using Insertion Loss Method #RFDesign #Microwave #RFcircuit](https://www.youtube.com/watch?v=eV_d3_0MwU4) — We The Engineers - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/rf_and_microwave_engineering/insertion-loss-lpf-microstrip - Whole chapter: https://www.bctengineer.com/md/semester_7/rf_and_microwave_engineering/chapter-6 ## Question Describe insertion loss method of microwave filter design. Illustrate an example of a passive LPF using μ-strip. ## Short answer Insertion loss method uses PLR to choose response, find N, prototype g, scale to L,C. Example Chebyshev 0.5 dB LPF at 2 GHz, 50 Ω: C1=C3=2.541 pF, L2=4.364 nH; stepped-impedance microstrip lengths 8.06 mm (20 Ω) and 6.48 mm (120 Ω). ## Answer In the **insertion loss method**, the filter response is defined by the **power loss ratio**: ```text PLR = P_available / P_delivered = 1 / (1 − |Γ(ω)|²) IL (dB) = 10 log PLR ``` A physically realisable PLR must be a ratio of even polynomials in ω, so the designer chooses a polynomial that gives the wanted response. #### Steps of the method 1. **Specify:** fc, R0, ripple and minimum stopband attenuation. 2. **Choose the response:** Butterworth PLR = 1 + k²(ω/ωc)^(2N) for a flat passband; Chebyshev PLR = 1 + k²TN²(ω/ωc) for a sharper cut-off with ripple; linear-phase for flat group delay. 3. **Find N** from the stopband requirement. 4. **Prototype values** g0 … gN+1 from tables (R0 = 1 Ω, ωc = 1 rad/s). 5. **Scale** impedance and frequency: L = gR0/ωc, C = g/(R0ωc); transform to HP/BP/BS if needed. 6. **Realise** in transmission lines (Richards + Kuroda, stepped impedance), then simulate and tune. #### Example: passive LPF using microstrip Specs: fc = 2 GHz, R0 = 50 Ω, Chebyshev 0.5 dB ripple, N = 3 (π type: shunt C – series L – shunt C). ```text g1 = g3 = 1.5963, g2 = 1.0967, g4 = 1.0 C1 = C3 = 1.5963/(50×2π×2×10⁹) = 2.541 pF L2 = 1.0967×50/(2π×2×10⁹) = 4.364 nH ``` Stepped-impedance microstrip (εr = 4.2, h = 1.58 mm, Zℓ = 20 Ω, Zh = 120 Ω): ```text C: βl = g·Zℓ/R0 L: βl = g·R0/Zh ``` | Element | Line | βl | Length | Width | |---|---|---|---|---| | C1 | 20 Ω | 36.59° | 8.06 mm | 11.27 mm | | L2 | 120 Ω | 26.18° | 6.48 mm | 0.43 mm | | C3 | 20 Ω | 36.59° | 8.06 mm | 11.27 mm | ```text 50Ω ______ ______ 50Ω ======| |------| |====== | C1 | L2 | C3 | |______| thin |______| wide wide ``` All lines are shorter than λ/8 (45°), so the lumped approximation holds well. The 50 Ω feed lines are 3.13 mm wide. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the power loss ratio in the insertion loss method? A. PLR = 1 − |Γ|^2 B. PLR = P_delivered / P_available C. PLR = 1/(1 − |Γ(ω)|^2) = P_available / P_delivered D. PLR = |S21| Answer: C. PLR is available power over power delivered to the load, equal to 1/(1 − |Γ|^2), and IL = 10 log PLR. (Key point: Power loss ratio) 2. Which response gives a sharper cut-off at the cost of passband ripple? A. Butterworth B. Linear phase C. Chebyshev D. Maximally flat Answer: C. Chebyshev PLR = 1 + k²TN²(ω/ωc) has equal-ripple passband and sharper cut-off. (Key point: Choosing response) 3. For the 0.5 dB Chebyshev N = 3 LPF at 2 GHz, 50 Ω (g1 = g3 = 1.5963, g2 = 1.0967), what are C1 and L2? A. C1 = 2.541 pF, L2 = 4.364 nH B. C1 = 4.364 pF, L2 = 2.541 nH C. C1 = 1.59 pF, L2 = 7.96 nH D. C1 = 2.541 nH, L2 = 4.364 pF Answer: A. C1 = 1.5963/(50·2π·2e9) = 2.541 pF and L2 = 1.0967·50/(2π·2e9) = 4.364 nH. (Key point: Example element values) 4. In the stepped-impedance microstrip LPF, how are capacitors and inductors realised? A. Capacitors by narrow high-Z lines, inductors by wide low-Z lines B. Both by 50 Ω lines C. Capacitors by wide low-Z (20 Ω) lines, inductors by narrow high-Z (120 Ω) lines D. Capacitors by shorted stubs Answer: C. Wide low-impedance sections are capacitive and narrow high-impedance sections are inductive; lines shorter than λ/8 keep the lumped approximation valid. 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